6.3 Expected Value of a Discrete Random Variable

练习题 / Exercises

基础期望值计算 / Basic Expected Value Calculations

问题 1 / Question 1

随机变量X具有以下概率分布，求E(X)：

The random variable X has the following probability distribution. Find E(X):

x	-1	0	1	2	3
P(X = x)	$\frac{1}{5}$	$\frac{1}{5}$	$\frac{1}{5}$	$\frac{1}{5}$	$\frac{1}{5}$

a) 求E(X)

a) Find E(X)

b) 求Var(X)

b) Find Var(X)

问题 2 / Question 2

随机变量X具有以下概率分布：

The random variable X has the following probability distribution:

x	1	2	3
P(X = x)	$\frac{1}{3}$	$\frac{1}{2}$	$\frac{1}{6}$

求E(X)和E(X²)。

Find E(X) and E(X²).

未知参数求解 / Finding Unknown Parameters

问题 3 / Question 3

随机变量X具有以下概率分布：

The random variable X has a probability distribution as shown in the table.

x	1	2	3	4	5
P(X = x)	0.1	p	0.3	q	0.2

给定E(X) = 3，求p和q的值。

Given that E(X) = 3, find the value of p and the value of q.

问题 4 / Question 4

随机变量X具有概率函数：

The random variable X has a probability function:

\[P(X = x) = \frac{1}{x}, \quad x = 2, 3, 6\]

a) 构造X和X²的概率分布表。

a) Construct tables giving the probability distributions of X and X².

b) 求E(X)和E(X²)。

b) Work out E(X) and E(X²).

c) 判断[E(X)]² = E(X²)是否成立。

c) State whether or not [E(X)]² = E(X²).

实际应用问题 / Practical Application Problems

问题 5 / Question 5

一个公司生产手机壳。每50个手机壳中有一个是次品，但公司在买家投诉前不知道哪些是次品。假设公司销售正常手机壳获利3美元，但次品手机壳需要更换，造成8美元损失。

A company makes phone covers. One out of every 50 phone covers is faulty, but the company doesn't know which ones are faulty until a buyer complains. Suppose the company makes a $3 profit on the sale of any working phone cover, but suffers a loss of $8 for every faulty phone cover due to replacement costs.

计算每个手机壳的期望利润，无论是否为次品。

Calculate the expected profit for each phone cover, regardless of whether or not it is faulty.

问题 6 / Question 6

三枚公平的六面骰子被掷出。离散随机变量X定义为三个值中最大的值。求E(X)。

Three fair six-sided dice are rolled. The discrete random variable X is defined as the largest value of the three values shown. Find E(X).

挑战题 / Challenge Problems

问题 7 / Question 7

随机变量X具有概率函数：

The random variable X has a probability function:

\[P(X = x) = \frac{2}{x^2}, \quad x = 2, 3, 4\]

a) 解释为什么这个函数不描述概率分布。

a) Explain how you know that Marie's function does not describe a probability distribution.

b) 给定正确的概率函数形式为：

b) Given that the correct probability function is in the form:

\[P(X = x) = \frac{k}{x^2}, \quad x = 2, 3, 4\]

其中k是常数，求k的准确值。

where k is a constant, find the exact value of k.

答案与解析 / Answers and Solutions

问题 1 答案 / Answer to Question 1

a) E(X) = Σx P(X = x) = (-1)(1/5) + 0(1/5) + 1(1/5) + 2(1/5) + 3(1/5) = 1

b) 首先求E(X²) = Σx² P(X = x) = 1(1/5) + 0(1/5) + 1(1/5) + 4(1/5) + 9(1/5) = 3

b) First find E(X²) = Σx² P(X = x) = 1(1/5) + 0(1/5) + 1(1/5) + 4(1/5) + 9(1/5) = 3

Var(X) = E(X²) - [E(X)]² = 3 - 1² = 2

问题 2 答案 / Answer to Question 2

E(X) = 1×(1/3) + 2×(1/2) + 3×(1/6) = 1/3 + 1 + 1/2 = 1.833...

E(X²) = 1×(1/3) + 4×(1/2) + 9×(1/6) = 1/3 + 2 + 1.5 = 3.833...

问题 3 答案 / Answer to Question 3

概率总和：0.1 + p + 0.3 + q + 0.2 = 1 → p + q = 0.4

Sum of probabilities: 0.1 + p + 0.3 + q + 0.2 = 1 → p + q = 0.4

期望值：1×0.1 + 2×p + 3×0.3 + 4×q + 5×0.2 = 3

Expected value: 1×0.1 + 2×p + 3×0.3 + 4×q + 5×0.2 = 3

0.1 + 2p + 0.9 + 4q + 1 = 3

2p + 4q + 2 = 3

2p + 4q = 1

将第一个方程乘以2：2p + 2q = 0.8

Multiply first equation by 2: 2p + 2q = 0.8

减去：(2p + 4q) - (2p + 2q) = 1 - 0.8

2q = 0.2 → q = 0.1

p + 0.1 = 0.4 → p = 0.3

问题 4 答案 / Answer to Question 4

a) X的概率分布：

a) Probability distribution of X:

x	2	3	6
P(X = x)	1/2	1/3	1/6

X²的概率分布：

Probability distribution of X²:

x²	4	9	36
P(X² = x²)	1/2	1/3	1/6

b) E(X) = 2×(1/2) + 3×(1/3) + 6×(1/6) = 1 + 1 + 1 = 3

E(X²) = 4×(1/2) + 9×(1/3) + 36×(1/6) = 2 + 3 + 6 = 11

c) [E(X)]² = 9 ≠ 11 = E(X²)，不成立。

c) [E(X)]² = 9 ≠ 11 = E(X²), not equal.

问题 5 答案 / Answer to Question 5

正常手机壳的概率：49/50

Probability of working phone cover: 49/50

次品手机壳的概率：1/50

Probability of faulty phone cover: 1/50

期望利润 = 3 × (49/50) + (-8) × (1/50) = (147/50) - (8/50) = 139/50 = 2.78 美元

Expected profit = 3 × (49/50) + (-8) × (1/50) = (147/50) - (8/50) = 139/50 = $2.78

问题 6 答案 / Answer to Question 6

三枚骰子的最大值X的概率分布：

Probability distribution of X (maximum of three dice):

P(X=1) = (1/6)^3 = 1/216

P(X=2) = (2/6)^3 - (1/6)^3 = 8/216 - 1/216 = 7/216

P(X=3) = (3/6)^3 - (2/6)^3 = 27/216 - 8/216 = 19/216

P(X=4) = (4/6)^3 - (3/6)^3 = 64/216 - 27/216 = 37/216

P(X=5) = (5/6)^3 - (4/6)^3 = 125/216 - 64/216 = 61/216

P(X=6) = 1 - (5/6)^3 = 1 - 125/216 = 91/216

E(X) = 1×(1/216) + 2×(7/216) + 3×(19/216) + 4×(37/216) + 5×(61/216) + 6×(91/216)

= (1 + 14 + 57 + 148 + 305 + 546)/216 = 1071/216 = 4.958...

= (1 + 14 + 57 + 148 + 305 + 546)/216 = 1071/216 ≈ 4.958

问题 7 答案 / Answer to Question 7

a) 概率总和 = 2/4 + 2/9 + 2/16 = 0.5 + 0.222... + 0.125 = 0.847... ≠ 1

a) Sum of probabilities = 2/4 + 2/9 + 2/16 = 0.5 + 0.222... + 0.125 = 0.847... ≠ 1

b) k/4 + k/9 + k/16 = 1

k(1/4 + 1/9 + 1/16) = 1

k(9/36 + 4/36 + 9/36) = 1

k(22/36) = 1

k = 36/22 = 18/11